Car B travels a distance <em>x</em> after time <em>t</em> (measured in seconds) of
<em>x</em> = (80 m/s) <em>t</em>
so that at 6:40 AM (which is 40 min = 2400 s after the starting time of 6:00 AM), it will have moved
(80 m/s) * (2400 s) = 192,000 m
If we take <em>t</em> = 0 to refer to 6:40 AM, the cars' respective positions <em>x</em> at time <em>t</em> are given by
A: <em>x</em> = (280 m/s) <em>t</em>
B: <em>x</em> = 192,000 m + (80 m/s) <em>t</em>
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(i) After 1 h = 3600 s, their positions are
A: (280 m/s) * (3600 s) = 1,008,000 m
B: 192,000 m + (80 m/s) * (3600 s) = 480,000 m
away from the point X, so the distance between the two cars is
1,008,000 m - 480,000 m = 528,000 m = 582 km
(ii) Car A moves faster than car B, so A overtakes B as soon as their positions are equal, which happens for
(280 m/s) <em>t</em> = 192,000 m + (80 m/s) <em>t</em>
(200 m/s) <em>t</em> = 192,000 m
<em>t</em> = (192,000 m) / (200 m/s)
<em>t</em> = 960 s = 16 min
So after 16 min, both cars have a position of
(280 m/s) * (960 s) = 268,800 m = 268.8 km
beyond point X.